Canonical products and general form for entire functions
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In Ahlfors book, chapter V - 2.3, we have a method to find a form to entire functions, that is: if $f$ is an entire function with a finite number of zeros $a_1, ..., a_N$, and a zero of multiplicity $m$ on the origin then $$f(z)=z^m e^{g(z)}prod_{n = 1}^{N} (1-frac{z}{a_n})$$
after, generalyze this for infinitely many zeros
$$f(z)=z^m e^{g(z)}prod_{n = 1}^{infty} (1-frac{z}{a_n})$$
My question is: Why that generalization doesn't mean $f$ is a constant function? I said that because we know that an analytic function (nonconstant) had a finite number of zeros. My point is a contradiction between this two facts.
complex-analysis entire-functions
asked Dec 14 '18 at 19:58
Júlio César M. MarquesJúlio César M. Marques
616
$endgroup$
add a comment |
$begingroup$
In Ahlfors book, chapter V - 2.3, we have a method to find a form to entire functions, that is: if $f$ is an entire function with a finite number of zeros $a_1, ..., a_N$, and a zero of multiplicity $m$ on the origin then $$f(z)=z^m e^{g(z)}prod_{n = 1}^{N} (1-frac{z}{a_n})$$
after, generalyze this for infinitely many zeros
$$f(z)=z^m e^{g(z)}prod_{n = 1}^{infty} (1-frac{z}{a_n})$$
My question is: Why that generalization doesn't mean $f$ is a constant function? I said that because we know that an analytic function (nonconstant) had a finite number of zeros. My point is a contradiction between this two facts.
complex-analysis entire-functions
asked Dec 14 '18 at 19:58
Júlio César M. MarquesJúlio César M. Marques
616
$endgroup$
1
$begingroup$
Doesn't the sine function have infinitely many zeroes?
$endgroup$
– Lord Shark the Unknown
Dec 14 '18 at 20:01
add a comment |
$begingroup$
In Ahlfors book, chapter V - 2.3, we have a method to find a form to entire functions, that is: if $f$ is an entire function with a finite number of zeros $a_1, ..., a_N$, and a zero of multiplicity $m$ on the origin then $$f(z)=z^m e^{g(z)}prod_{n = 1}^{N} (1-frac{z}{a_n})$$
after, generalyze this for infinitely many zeros
$$f(z)=z^m e^{g(z)}prod_{n = 1}^{infty} (1-frac{z}{a_n})$$
My question is: Why that generalization doesn't mean $f$ is a constant function? I said that because we know that an analytic function (nonconstant) had a finite number of zeros. My point is a contradiction between this two facts.
complex-analysis entire-functions
asked Dec 14 '18 at 19:58
Júlio César M. MarquesJúlio César M. Marques
616
$endgroup$
In Ahlfors book, chapter V - 2.3, we have a method to find a form to entire functions, that is: if $f$ is an entire function with a finite number of zeros $a_1, ..., a_N$, and a zero of multiplicity $m$ on the origin then $$f(z)=z^m e^{g(z)}prod_{n = 1}^{N} (1-frac{z}{a_n})$$
after, generalyze this for infinitely many zeros
$$f(z)=z^m e^{g(z)}prod_{n = 1}^{infty} (1-frac{z}{a_n})$$
My question is: Why that generalization doesn't mean $f$ is a constant function? I said that because we know that an analytic function (nonconstant) had a finite number of zeros. My point is a contradiction between this two facts.
complex-analysis entire-functions
complex-analysis entire-functions
asked Dec 14 '18 at 19:58
Júlio César M. MarquesJúlio César M. Marques
616
asked Dec 14 '18 at 19:58
Júlio César M. MarquesJúlio César M. Marques
616
asked Dec 14 '18 at 19:58
Júlio César M. MarquesJúlio César M. Marques
616
asked Dec 14 '18 at 19:58
Júlio César M. MarquesJúlio César M. Marques
616
asked Dec 14 '18 at 19:58
Júlio César M. MarquesJúlio César M. Marques
616
616
1
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Doesn't the sine function have infinitely many zeroes?
$endgroup$
– Lord Shark the Unknown
Dec 14 '18 at 20:01
add a comment |
1
$begingroup$
Doesn't the sine function have infinitely many zeroes?
$endgroup$
– Lord Shark the Unknown
Dec 14 '18 at 20:01
1
1
$begingroup$
Doesn't the sine function have infinitely many zeroes?
$endgroup$
– Lord Shark the Unknown
Dec 14 '18 at 20:01
$begingroup$
Doesn't the sine function have infinitely many zeroes?
$endgroup$
– Lord Shark the Unknown
Dec 14 '18 at 20:01
add a comment |
1 Answer
1
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oldest
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$begingroup$
You should mention that in the finite case multiple zeros are repeated. Ahlfors says that if there are infinitely many zeros one can try to obtain a similar representation by means of an infinite product. He states that this product converges absolutely if and only if $sum_{n=1}^infty 1/lvert a_n rvert$ is convergent.
A neccessary condition for the convergence of this series is that $a_n to infty$ as $n to infty$. This means that $(a_n)$ does not have a cluster point in $mathbb{C}$. Therefore you cannot conclude that $f$ is constant. See Lord Shark the Unknown's comment.
answered Dec 14 '18 at 22:43
Paul FrostPaul Frost
10.7k3933
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add a comment |
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$begingroup$
You should mention that in the finite case multiple zeros are repeated. Ahlfors says that if there are infinitely many zeros one can try to obtain a similar representation by means of an infinite product. He states that this product converges absolutely if and only if $sum_{n=1}^infty 1/lvert a_n rvert$ is convergent.
A neccessary condition for the convergence of this series is that $a_n to infty$ as $n to infty$. This means that $(a_n)$ does not have a cluster point in $mathbb{C}$. Therefore you cannot conclude that $f$ is constant. See Lord Shark the Unknown's comment.
answered Dec 14 '18 at 22:43
Paul FrostPaul Frost
10.7k3933
$endgroup$
add a comment |
$begingroup$
You should mention that in the finite case multiple zeros are repeated. Ahlfors says that if there are infinitely many zeros one can try to obtain a similar representation by means of an infinite product. He states that this product converges absolutely if and only if $sum_{n=1}^infty 1/lvert a_n rvert$ is convergent.
A neccessary condition for the convergence of this series is that $a_n to infty$ as $n to infty$. This means that $(a_n)$ does not have a cluster point in $mathbb{C}$. Therefore you cannot conclude that $f$ is constant. See Lord Shark the Unknown's comment.
answered Dec 14 '18 at 22:43
Paul FrostPaul Frost
10.7k3933
$endgroup$
add a comment |
$begingroup$
You should mention that in the finite case multiple zeros are repeated. Ahlfors says that if there are infinitely many zeros one can try to obtain a similar representation by means of an infinite product. He states that this product converges absolutely if and only if $sum_{n=1}^infty 1/lvert a_n rvert$ is convergent.
A neccessary condition for the convergence of this series is that $a_n to infty$ as $n to infty$. This means that $(a_n)$ does not have a cluster point in $mathbb{C}$. Therefore you cannot conclude that $f$ is constant. See Lord Shark the Unknown's comment.
answered Dec 14 '18 at 22:43
Paul FrostPaul Frost
10.7k3933
$endgroup$
You should mention that in the finite case multiple zeros are repeated. Ahlfors says that if there are infinitely many zeros one can try to obtain a similar representation by means of an infinite product. He states that this product converges absolutely if and only if $sum_{n=1}^infty 1/lvert a_n rvert$ is convergent.
A neccessary condition for the convergence of this series is that $a_n to infty$ as $n to infty$. This means that $(a_n)$ does not have a cluster point in $mathbb{C}$. Therefore you cannot conclude that $f$ is constant. See Lord Shark the Unknown's comment.
answered Dec 14 '18 at 22:43
Paul FrostPaul Frost
10.7k3933
answered Dec 14 '18 at 22:43
Paul FrostPaul Frost
10.7k3933
answered Dec 14 '18 at 22:43
Paul FrostPaul Frost
10.7k3933
answered Dec 14 '18 at 22:43
Paul FrostPaul Frost
10.7k3933
10.7k3933
add a comment |
add a comment |
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$begingroup$
Doesn't the sine function have infinitely many zeroes?
$endgroup$
– Lord Shark the Unknown
Dec 14 '18 at 20:01